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As solved in the video under 5 minutes, this is the notes for the 5 MCQs solved with explanation. MCQ type questions are not only helpful for your examinations but also for competitive exams like GATE, UPSC and many more.

Consider two forces $P$ and $Q$ acting simultaneously on a particle $O$ as shown in the figure above. Let the angle between these two forces be $\theta$. By extending imaginary lines from $P$ and $Q$, we can get a parallelogram as shown in the figure above as dotted lines.

Therefore, by the Law of Parallelogram of Forces the resultant of the forces $P$ and $Q$ can be represented by the diagonal of this parallelogram and its direction as $\alpha \degree$ from force $P$.

If two forces acting simultaneously on a particle are represented in magnitude and direction by the two sides of a parallelogram, then their resultant can be represented in **magnitude and direction by the diagonal of the parallelogram** which passes through their point of intersection.

And the magnitude of the resultant is given by:

$R = \sqrt{P^2 + Q^2 + 2 \ PQ \ \cos \theta}$And the direction of the resultant is given by:

$\tan \alpha = {Q \ \sin \theta \over P + Q \ \cos \theta}$Now, let's start solving MCQs. These questions will have 2 forces acting on a particle, hence we will apply the law of parallelogram of forces.

a) $P+Q$ c) $P \over Q$

b) $P-Q$ d) $Q \over P$

Given:

$\theta = 180 \degree$ (because P and Q are in straight line, but opposite in direction)

a) $2P \sin {\theta \over 2}$ c) $2P \tan {\theta \over 2}$

b) $2P \cos {\theta \over 2}$ d) $2P \cot {\theta \over 2}$

Given:

$P = Q$

a) $0 \degree \And 180 \degree$ c) $90 \degree \And 180 \degree$

b) $180 \degree \And 0 \degree$ d) $90 \degree \And 0 \degree$

__Explanation:__

In the question it is asked to find the value of $\theta$ for the maximum value $R$ and the minimum value of $R$.

We know that: $R = \sqrt{P^2 + Q^2 + 2 \ PQ \ \cos \theta}$.

Now if we change the value of $cos \theta$ and add or subtract it from the remaining value of $P$ and $Q$ we get the value of $R$.

So, to find the answer, let's take 4 extreme values from the unit circle as given in the options for $\theta \to 0 \degree, \ 90 \degree, \ 180 \degree \ and \ 270 \degree$

Now the $\cos$ values for these angles will be $\to 1, 0, -1 \ and \ 0$ respectively.

Here, maximum value is $1$(i.e. $\cos 0$) and minimum value is $-1$(i.e. $\cos 180$).

Therefore, at $0 \degree$ and at $180 \degree$, the resultant will be maximum and minimum respectively.

a) ${P \over 2}$ c) ${P \over 2 \sqrt2}$

b) ${P \over \sqrt2}$ d) $\sqrt{2} \sdot P$

Given:

$P = Q$, $\theta = 90\degree$

a) ${P = Q}$ c) ${Q = R}$

b) ${Q = 2R}$ d) $\text{none of these}$

__Explanation:__

We have two set of data here:

- Forces $P$ and $Q$, whose resultant is R
- If $Q$ is doubled, then resultant(R') would be perpendicular to $P$. This means $\alpha = 90 \degree$ and $Q = 2Q$

This is only possible if:

$\begin{aligned} &P + Q \ \cos \theta = 0 \\ &\boxed{\therefore P = -2Q \cos \theta} ---- (1) \end{aligned}$Using the formula for resultant force:

$\begin{aligned} R &= \sqrt{P^2 + Q^2 + 2 \ PQ \ \cos \theta} \\ &\text{Using from equation (1)} \\ &= \sqrt{(-2Q \cos \theta)^2 + Q^2 + 2(-2Q \cos \theta) \sdot Q cos \theta} \\ &= \sqrt{4Q^2 \cos^2 \theta + Q^2 - 4Q^2 \cos^2 \theta} \\ &= \sqrt{Q^2} \\ &\boxed{\therefore R = Q} \end{aligned}$- Balancing of V-Engines
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